Recent content by hedegaardo1

Hm okay. So since 1 mol P is proportional to 2 moles C and we go from P to C the conversion factor is 2 mol C / 1 mol P, that is [C] = 2 [P] ?

That was what I tried in the original post. I mean we have P --> 2C+D. So the concentration of C is 2/3 of the original concentration, that is [C]=2/3 * [P]?

Well I have the instantaneous reaction rates $$ -\frac{d[P]}{dt}=\frac{1}{2}\frac{d[C]}{dt}$$ which gives $$k'[P]^2=\frac{1}{2}\frac{d[C]}{dt}\Rightarrow \int_0^t k[P]^2 dt = \frac{1}{2}\int_0^{[C]}d[C]\Rightarrow k[P]^2 t = \frac{[C]}{2}$$ but if it is instantaneous t --> 0 and the above is...

I'm really not sure what you want me to realize by this. Then $$-\frac{d[P]}{dt}= k' [P]^2$$ but I don't see how this helps me determine the concentration of [C].

Homework Statement The second-order rate constant for the reaction A+2B --> 2C+D is 0.34 dm3/mol s. What is the concentration of C after 20 s and 15 min if the initial concentrations were [A] = 0.027 mol/dm3 and [ B] = 0.130 mol/dm3 Homework Equations The integrated rate law for A+2B --> P...

I'm trying to solve the same problem and got to the same conclusion. But how do you justify that the surface integral is zero?